What is the vertex of $y = - 3 x^{2} - 4 x + 2 {(x - 2)}^{2}$ $y=-3x^2-4x+2(x-2)^2$ ?

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What is the vertex of $y = - 3 x^{2} - 4 x + 2 {(x - 2)}^{2}$ $y=-3x^2-4x+2(x-2)^2$ ?

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Answer 1 · 2015-12-23T07:22:57

(4,24)

Explanation:

Simplify first
$y = - 3 x^{2} - 4 x + 2 {(x - 2)}^{2}$ $y= -3x^2-4x+2(x-2)^2$
$y = - 3 x^{2} - 4 x + 2 (x^{2} + 4 x + 4)$ $y= -3x^2 -4x + 2(x^2+4x+4)$
$y = - 3 x^{2} - 4 x + 2 x^{2} + 8 x + 8$ $y= -3x^2 -4x + 2x^2 +8x+8$
$y = - x^{2} + 8 x + 8$ $y= -x^2 +8x+8$

Now to solve for the vertex algebraically , we use the formula
Vertex= $(- \frac{b}{2 a}, f (- \frac{b}{2 a}))$ $( -b/(2a) , f(-b/(2a) ) )$
$- \frac{b}{2 a} = 4$ $-b/(2a) = 4$
$f (4) = 24$ $f(4)= 24$

Vertex = (4,24)

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What is the vertex of $y = - 3 x^{2} - 4 x + 2 {(x - 2)}^{2}$ $y=-3x^2-4x+2(x-2)^2$ ?

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What is the vertex of $y = - 3 x^{2} - 4 x + 2 {(x - 2)}^{2}$ $y=-3x^2-4x+2(x-2)^2$ ?